$\epsilon - \delta$ proof of $\lim _{x \rightarrow 0} \frac{2 x+3}{x+1}=3$

Question

A hint is given: $|x| < 1/2 \implies |x+1| > 1/2$.

I have been advised to only use the general method. which is starting from

$$\left|\frac{2 x+3}{x+1}-3\right|$$

and ending with something with $|x-0|$.

I have come this far

$$ \left|\frac{2 x+3}{x+1}-3\right|=\left|\frac{-x}{x+1}\right|=\frac{|x|}{|x+1|} $$

I think that I have to use the hint now, but I can't figure out how and with what purpose.

Answer 1

The way $\delta-\epsilon$ proofs is that for ever $\epsilon >0$ there is a $\delta >0$ so that $|x-a| < \delta \implies |f(x)-L| < \epsilon$.

We have two possible strategies. Either start at the end and work back:

$|f(x) - L| <\epsilon \Leftarrow$

....

$|x - a| < h(\epsilon) = \delta$

Or

Start at the beginning and work forward.

$|x-a| < \delta \implies$

....

$|f(x) - L| < j(\delta) \le \epsilon$ and we work backwards to figure out what $\delta$ is in terms of $\epsilon$.

...

In this case the hint is: If $|x| < \frac 12$ then $|x+1| = x+1 >\frac 12$ and that $\frac 1{|x+1|} < 2$.

So if we choose our $\delta \le \frac 12$ then we can replace any occurance of $\frac M{|x+1|}$ with $\frac M{|x+1|} < \frac M2$.

That makes this work, whether front to back or back to front short work:

If $|x| < \delta \le \frac 12$ then

$2{|x|}2 < 2\delta $

$\frac {|x|}{|x+1|} < \frac {|x|}{\frac 12}=2|x| \le 2 \delta $ and

$|\frac {2x+1}{x+1} -3|=|\frac {-x}{x+1}| = \frac {|x|}{|x+1|}<2{|x|} \le 2 \delta $.

So as long as we choose $\delta$ so that $2\delta \le \epsilon$ and $\delta < \frac 12$, or in other words if $\delta \le \min(\frac 12, \frac\epsilon 2)$ we will get our result.

Or going the other way:

$|\frac {2x+1}{x+1} - 3| = |\frac {-x}{x+1}|=\frac {|x|}{|x+1|}< \epsilon$

Is implied by $\frac{|x|}{|x+1|} < 2{|x|} < \epsilon$ if $|x| < \frac 12$

Which would follow if $|x| < \frac{\epsilon}2$ and $|x| < \frac 12$.

So if $\delta = \min (\frac\epsilon 2, \frac 12)$ we'll be fine.

Answer 2

It follows from that hint that$$\lvert x\vert<\frac12\implies\left\lvert\frac{2x+3}{x+1}-3\right\rvert<2\lvert x\rvert.$$So, take $\delta=\min\left\{\frac\varepsilon2,\frac12\right\}$. Then, if $\lvert x\rvert<\delta$, then$$\left\lvert\frac{2x+3}{x+1}-3\right\rvert<2\lvert x\rvert.\tag1$$And now, since $\lvert x\rvert<\frac\varepsilon2$, you deduce from $(1)$ that$$\left\lvert\frac{2x+3}{x+1}-3\right\rvert<\varepsilon.$$